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course Phy 122
2/24 9:26This is the revision of the Bottle Thermometer Lab.
My corrections are set off with &&&&" "
bottle thermometer
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Phy 122
Your 'bottle thermometer' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Bottle Thermometer_labelMessages **
2/17 9pm
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About 5 hours
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You can use the bottle, stopper and tubes as a very sensitive thermometer. This thermometer will have excellent precision, clearly registering temperature changes on the order of .01 degree. The system will also demonstrate a very basic thermal engine and its thermodynamic properties.
Set up your system with a vertical tube and a pressure-indicating tube, as in the experiment on measuring atmospheric pressure. There should be half a liter or so of water in the bottom of the container.
Refer back to the experiment 'Measuring Atmospheric Pressure' for a detailed description of how the pressure-indicating tube is constructed for the 'stopper' version of the experiment.
For the bottle-cap version, the pressure-indicating tube is the second-longest tube. The end inside the bottle should be open to the gas inside the bottle (a few cm of tube inside the bottle is sufficient) and the other end should be capped.
The figure below shows the basic shape of the tube; the left end extends down into the bottle and the capped end will be somewhere off to the right. The essential property of the tube is that when the pressure in the bottle increases, more force is exerted on the left-hand side of the 'plug' of liquid, which moves to the right until the compression of air in the 'plugged' end balances it. As long as the liquid 'plug' cannot 'leak' its liquid to the left or to the right, and as long as the air column in the plugged end is of significant length so it can be measured accurately, the tube is set up correctly.
If you pressurize the gas inside the tube, water will rise accordingly in the vertical tube. If the temperature changes but the system is not otherwise tampered with, the pressure and hence the level of water in the tube will change accordingly.
When the tube is sealed, pressure is atmospheric and the system is unable to sustain a water column in the vertical tube. So the pressure must be increased. Various means exist for increasing the pressure in the system.
You could squeeze the bottle and maintain enough pressure to support, for example, a 50 cm column. However the strength of your squeeze would vary over time and the height of the water column would end up varying in response to many factors not directly related to small temperature changes.
You could compress the bottle using mechanical means, such as a clamp. This could work well for a flexible bottle such as the one you are using, but would not generalize to a typical rigid container.
You could use a source of compressed air to pressurize the bottle. For the purposes of this experiment, a low pressure, on the order of a few thousand Pascals (a few hudredths of an atmosphere) would suffice.
The means we will choose is the low-pressure source, which is readily available to every living land animal. We all need to regularly, several times a minute, increase and decrease the pressure in our lungs in order to breathe. We're going to take advantage of this capacity and simply blow a little air into the bottle.
Caution: The pressure you will need to exert and the amount of air you will need to blow into the system will both be less than that required to blow up a typical toy balloon. However, if you have a physical condition that makes it inadvisable for you to do this, let the instructor know. There is an alternative way to pressurize the system.
You recall that it takes a pretty good squeeze to raise air 50 cm in the bottle. You will be surprised at how much easier it is to use your diaphragm to accomplish the same thing. If you open the 'pressure valve', which in this case consists of removing the terminating cap from the third tube, you can then use the vertical tube as a 'drinking straw' to draw water up into it. Most people can easily manage a 50 cm; however don't take this as a challenge. This isn't a test of how far you can raise the water.
Instructions follow:
Before you put your mouth on the tube, make sure it's clean and make sure there's nothing in the bottle you wouldn't want to drink. The bottle and the end of the tube can be cleaned, and you can run a cleaner through the tube (rubbing alcohol works well to sterilize the tube). If you're careful you aren't likely to ingest anything, but of course you want the end of the tube to be clean.
Once the system is clean, just do this. Pull water up into the tube. While maintaining the water at a certain height, replace the cap on the pressure-valve tube and think for a minute about what's going to happen when you remove the tube from your mouth. Also think about what, if anything, is going to happen to the length of the air column at the end of the pressure-indicating tube. Then go ahead and remove the tube from your mouth and watch what happens.
Describe below what happens and what you expected to happen. Also indicate why you think this happens.
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I expected that the water in the verical tube would stay put at the level to which I had drawn it. I expected that the length of the air colum in the pressure indicating tube would lengthen as I drew up the water and return to its origianl position after I removed my mouth.
The water in the verical tube fell (unexpectedly) from the height I had drawn it to a height about 10 cm about the level of the water in the bottle before slowly falling to the level of the water. I'm not sure why this happened. The air column at the end of the pressure indicating tube got longer because the as the water effectively left the bottle and entered the vertical tube, the volume available for air at the same pressure in the bottle increased, so some of the air in the tube flowed back into the bottle bringing the water plug slightly back towards the bottle. When I removed my mouth and the water in the vertical tube fell, the air in the bottle was forced back up into the pressure indicating tube.
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Now think about what will happen if you remove the cap from the pressure-valve tube. Will air escape from the system? Why would you or would you not expect it to do so?
Go ahead and remove the cap, and report your expectations and your observations below.
****
Some air might escape the system but I believe at this point the pressure in the bottle is nearly equal to the pressure outside the bottle, so there is no sound or feel to make me think there is any great amount of escape going on.
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Now replace the cap on the pressure-valve tube and, while keeping an eye on the air column in the pressure-indicating tube, blow just a little air through the vertical tube, making some bubbles in the water inside the tube. Blow enough that the air column in the pressure-indicating tube moves a little, but not more than half a centimeter or so. Then remove the tube from your mouth, keeping an eye on the pressure-indicating tube and also on the vertical tube.
What happens?
****
When I blow into the tube, the air column in the pressure indicating tube shortens. When I remove my mouth the air column returns to approximately its original position. A column of water did shoot up the vertical tube and then slowly subsided. I imagine that since I forced air into the system, that then bubbled up into the bottle that this increased pressure in the bottle forced the water back up the vertical tube once I stopped blowing. Not sure why the water then slowly subsided.
???Where did that air go. I had adjusted the tubes in the cap and I did not hear much hissing this time, so I assumed my cap was a better seal this time around???
???I have found that very quickly in these experiments, my tubes become full of tiny plugs of water and tiny plugs of air. I tap and blow to clear them, but they only seem to multiply. Will this affect my results???
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Why did the length of the air column in the pressure-indicating tube change length when you blew air into the system? Did the air column move back to its original position when you removed the tube from your mouth? Did it move at all when you did so?
****
The air column shortened because I was increasing the amount of air in the system while maintaining the same volume capacity. This forced air into the pressure indicating tube, thus pushing on the water column. The air column did move back to roughly its same position when I removed the tube from my mouth.
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What happened in the vertical tube?
****
Water shot up the vertical column once I stopped blowing.
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Why did all these things happen? Which would would you have anticipated, and which would you not have anticipated?
****
A column of water did shoot up the vertical tube and then slowly subsided. I imagine that since I forced air into the system, that then bubbled up into the bottle that this increased pressure in the bottle forced the water back up the vertical tube once I stopped blowing. Not sure why the water then slowly subsided. Where did that air go?
I hadn't expected the water to shoot up the vetical column but now it makes sense why it did. the slow return done the tube doesn't make sense at this point.
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The slow return probably indicates a persistent leak, which allows the pressure to gradually equalize with the outside. It it's slow enough it won't seriously affect subsequent results. We'll see.
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What happened to the quantities P, V, n and T during various phases of this process?
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Pressure increased while blowing. volume remained constant. n remained constant for the water, I believe the n of air increased because I was forcing more air into the system. The temperture should have increased because the pressure increased.
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The n of the water is irrelevant; the P, Vn T and n that are related by the gas law all apply only to the gas. The water is just there to be moved around by changes in the gas.
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Place the thermometer that came with your kit near the bottle, with the bulb not touching any surface so that it is sure to measure the air temperature in the vicinity of the bottle and leave it alone until you need to read it.
Now you will blow enough air into the bottle to raise water in the vertical tube to a position a little ways above the top of the bottle.
Use the pressure-valve tube to equalize the pressure once more with atmospheric (i.e., take the cap off). Measure the length of the air column in the pressure-indicating tube, and as you did before place a measuring device in the vicinity of the meniscus in this tube.Replace the cap on the pressure-valve tube and again blow a little bit of air into the bottle through the vertical tube. Remove the tube from your mouth and see how far the water column rises. Blow in a little more air and remove the tube from your mouth. Repeat until water has reached a level about 10 cm above the top of the bottle.
Place the bottle in a pan, a bowl or a basin to catch the water you will soon pour over it.
Secure the vertical tube in a vertical or nearly-vertical position.
The water column is now supported by excess pressure in the bottle. This excess pressure is between a few hundredths and a tenth of an atmosphere.
The pressure in the bottle is probably in the range from 103 kPa to 110 kPa, depending on your altitude above sea level and on how high you chose to make the water column. You are going to make a few estimates, using 100 kPa as the approximate round-number pressure in the bottle, and 300 K as the approximate round-number air temperature. Using these ball-park figures:
If gas pressure in the bottle changed by 1%, by how many N/m^2 would it change?
****
If the pressure increased by 1% it would go up 1 kPa or 1000 Pa so therefore 1000 N/m^2
??? Note: I cannot get the water to stay at the level 10cm above the cap. Once I stop blowing, it slowly returns to the level of the water in the bottle. I thought about this for a while and figured I need to cap the vertical tube in order to keep the water in the column at a height of about 10cm above the cap. It seems to hold. I can't think of a way to keep the water up there unless I have the verical tube capped. Is this right???
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What would be the corresponding change in the height of the supported air column?
****
I would estimate a 1% increase in height.
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You can calculate the pressure necessary to support a water column of height 10 cm.
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If the pressure increased by 1% it would go up 1 kPa or 1000 Pa so therefore 1000 N/m^2
P = rho g h
1000N/m^2 = 1000kg/m^3 *9.8 m/s^ * h
h = 1000N/m^2 /(1000kg/m^3 *9.8 m/s^)
h = .10 m approximately
Likewise the pressure necessary to support a water column of 10 cm would be
P = rho g h
P = 1000kg/m^3 *9.8 m/s^ * .1m
P= 980 N/m^2 which is very close to the 1000N/m^2 we would get with a 1% increase in pressure.
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By what percent would air temperature have to change to result in this change in pressure, assuming that the container volume remains constant?
****
Air temp would have to increase by 1% as well
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Continuing the above assumptions:
How many degrees of temperature change would correspond to a 1% change in temperature?
****
1% of 300 degrees Kelvin would be 3 degrees Kelvin
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How much pressure change would correspond to a 1 degree change in temperature?
****
One degree would be a .33% increase in temp and therefore .33% increase in pressure or .33kPa increase
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By how much would the vertical position of the water column change with a 1 degree change in temperature?
****
Vertical position would increase .33%. Since my baseline positon is about 32 cm above the water level in the bottle, I would estimate the increase at about .1 cm.
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You need to find the change in the water column height that would require an additional 330 Pa of pressure for support.
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P = rho g h
330N/m^2 = 1000kg/m^3 *9.8 m/s^ * h
h = 330N/m^2 /(1000kg/m^3 *9.8 m/s^)
h = .034 m approximately
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How much temperature change would correspond to a 1 cm difference in the height of the column?
****
1 cm increase is about 3% increase in height. This would require a 3% increase in temp or approx 9 degrees
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The % change in pressure doesn't translate to a comparable % change in height. The reason is that the additional pressure is on top of the existing original pressure.
An analogy: The force exerted by a rubber band might be 0 when the rubber band's length is 8 cm. If stretched to 9 cm its tension might be 2 Newtons. It does not follow that at a 10 cm length its pressure will be 10/9 * 2 Newtons; the 11% increase in length does not result in an 11% increase in tension. Our expectation would be that since an extra cm of length between 8 and 9 cm added 2 N to the force, another cm would add about that much again.
See also my previous notes on how to relate the height of the column to the additional pressure.
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A 1 cm increase in height would require a specific increase in pressure
P = rho g h
P = 1000kg/m^3 *9.8 m/s^ * .01cm
P = 98 N/m^2
98 Pa/100 kPa = .00098 or a .098% increase.
An increase in pressure would be matched by the same percent increase in temp.
An increase of .098% of 300 Kelvin would be an increase of approximately .294 degrees Kelvin.
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How much temperature change would correspond to a 1 mm difference in the height of the column?
****
Temperature change would be about .9 degrees
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This depends on the correct relationship between pressure change and water column height.
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A 1mm difference in height would then mean a temperature increase of .0294 degrees.
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A change in temperature of 1 Kelvin or Celsius degree in the gas inside the container should correspond to a little more than a 3 cm change in the height of the water column. A change of 1 Fahrenheit degree should correspond to a little less than a 2 cm change in the height of the water column. Your results should be consistent with these figures; if not, keep the correct figures in mind as you make your observations.
The temperature in your room is not likely to be completely steady. You will first see whether this system reveals any temperature fluctuations:
Make a mark, or fasten a small piece of clear tape, at the position of the water column.
Observe, at 30-second intervals, the temperature on your alcohol thermometer and the height of the water column relative to the mark or tape (above the tape is positive, below the tape is negative).
Try to estimate the temperatures on the alcohol thermometer to the nearest .1 degree, though you won't be completely accurate at this level of precision.
Make these observations for 10 minutes.
Report in units of Celsius vs. cm your 20 water column position vs. temperature observations, in the form of a comma-delimited table below.
17.0, 0
17.0, 0
17.0, 0
17.0, 0
17.0, 0
17.0, 0
17.0, 0
17.0, 0
17.0, 0
17.0, 0
17.0, 0
17.0, 0
17.0, 0
17.0, 0
17.0, 0
17.0, 0
17.0, 0
17.0, 0
17.0, 0
17.0, 0
First column is temp in degress Celsius, second column is change in column position in cm
I noticed no flucuation at all in temp or column position over the course of 10 minutes
****
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Describe the trend of temperature fluctuations. Also include an estimate (or if you prefer two estimates) based on both the alcohol thermometer and the 'bottle thermometer' the maximum deviation in temperature over the 10-minute period. Explain the basis for your estimate(s):
****
I tried to find some temperature flucuations but the alchohol and bottle thermometers did not move at all during the 10 minutes. I tried to be very careful not to touch the thermometer or the bottle or even get too close to where I might affect the temperature near the thermometer with my own body temp.
I did observe that at 17 degrees is fairly cool in my basement to be standing in one position.
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I keep most of my house around 17 C in the winter, but wear an extra layer of flannel. It would be too cool otherwise.
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Now you will change the temperature of the gas in the system by a few degrees and observe the response of the vertical water column:
Read the alcohol thermometer once more and note the reading.
Pour a single cup of warm tap water over the sides of the bottle and note the water-column altitude relative to your tape, noting altitudes at 15-second intervals.
Continue until you are reasonably sure that the temperature of the system has returned to room temperature and any fluctuations in the column height are again just the result of fluctuations in room temperature. However don't take data on this part for more than 10 minutes.
Report your results below:
****
17.0
15, 2.2
30, 1.2
45, .3
60, .1
75, 0
90, 0
First row is temp in deg. Celsius according to the alchohol thermometer. Second through sixth rows are the elasped time in seconds, followed by change in water column position in cm.
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If your hands are cold, warm them for a minute in warm water. Then hold the palms of your hands very close to the walls of the container, being careful not to touch the walls. Keep your hands there for about a minute, and keep an eye on the air column.
Did your hands warm the air in the bottle measurably? If so, by how much? Give the basis for your answer:
****
My hands did warm the air measureabley. The water in the column rose about 2mm. My calculations for the change in water level due to change in temp were incorrect, but using the figures you provided, if 1 degree increase in temp (in Celsius) equals 3 cm of water column increase, a .2cm increase in water column height would correspond to a .07 degree increase in temperature.
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There's an advantage to the 17 C ambient temperature. In warm months the hands aren't that much colder than the surroundings and it's more difficult to observe a difference.
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Now reorient the vertical tube so that after rising out of the bottle the tube becomes horizontal. It's OK if some of the water in the tube leaks out during this process. What you want to achieve is an open horizontal tube,, about 30 cm above the level of water in the container, with the last few centimeters of the liquid in the horizontal portion of the tube and at least a foot of air between the meniscus and the end of the tube.
The system might look something like the picture below, but the tube running across the table would be more perfectly horizontal.
Place a piece of tape at the position of the vertical-tube meniscus (actually now the horizontal-tube meniscus). As you did earlier, observe the alcohol thermometer and the position of the meniscus at 30-second intervals, but this time for only 5 minutes. Report your results below in the same table format and using the same units you used previously:
****
I could not do this with an open ended horizontal tube. As soon as I uncapped the tube, the water column went back into the bottle. To continue the experiment, I changed the caps on my system (since it came with two caps) and this time I was able to hold the water column with the horizontal (formerly vertical) tube uncapped. With 3.5 hours already into the lab, I decided not to go back and re-do the earlier parts.
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I actually thought about advising you to reverse the caps, but wasn't close to a computer at the time so the advice never got sent. I'll need to put that into the writeup of the experiment, since students do very occasionally experience leaks.
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Repeat the experiment with your warm hands near the bottle. Report below what you observe:
****
I was able to move the water column about 70 cm before I had to step away to prevent the water from coming out of the tube and getting all over the floor (at this point finding a way to keep the tube as horizonatal as possible has precluded any good way to catch the water.
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When in the first bottle experiment you squeezed water into a horizontal section of the tube, how much additional pressure was required to move water along the horizontal section?
By how much do you think the pressure in the bottle changed as the water moved along the horizontal tube?
****
In the first bottle experiment, it did not take much additional pressure to move water along the horizontal section.
For this experiment, I think its hard to say how much the pressure changed in the bottle, because it doesn't take much at all in the way of additional pressure to move the water in a horiontal column. I say this also because I found it extremely hard to keep the tube horizontal
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If the water moved 10 cm along the horizontal tube, whose inner diameter is about 3 millimeters, by how much would the volume of air inside the system change?
****
With a diameter of about .3 cm this means the cross sectional area would be about .942 cm^2 so over 10cm this would be 9.42 cm^3 or 9.42mL.
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Diameter .3 cm implies radius .15 cm, and area pi r^2 = pi * (.15 cm)^2 = .07 cm^2, approximately. So it takes closer to 14 cm of tube to hold a cm^3 of water.
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By what percent would the volume of the air inside the container therefore change?
****
At this point I have lost so much water from mishaps, that I really don't know how much water is in the bottle. Since the initial instructions were to use about half a liter and not exactly half a liter, I did not take the time to carefully measure out the intital volume of water. Let's say though that I have .45 liters of water in my 2 liter bottle. if the air volume was 1.55 liters and now has an additional .00942 Liters, this would mean a percent change of .61%.
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Assuming constant pressure, how much change in temperature would be required to achieve this change in volume?
****
This would require a change in temp also of .61%. If air temp was originally 300K it would be a change of 1.83 degrees
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If the air temperature inside the bottle was 600 K rather than about 300 K, how would your answer to the preceding question change?
****
The temperature change would be twice as much or 3.66 degrees
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There were also changes in volume when the water was rising and falling in the vertical tube. Why didn't we worry about the volume change of the air in that case? Would that have made a significant difference in our estimates of temperature change?
****
I'm really not sure why it didn't matter in that case.
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The volume of the tube is only a few cm^3. Out of 1.5 liters, that's only about 1 part in 500. None of our observations are accurate to that level, so the corresponding volume changes are insignificant in the context of these observations.
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If the tube was not completely horizontal, would that affect our estimate of the temperature difference?
For example consider the tube in the picture below.
Suppose that in the process of moving 10 cm along the tube, the meniscus moves 6 cm in the vertical direction.
By how much would the pressure of the gas have to change to increase the altitude of the water by 6 cm?
****
I had to work backwards from the subsequent question. If the temperature when up 2 degrees or .67 percent, than so would the pressure go up .67 percent. At an original 100kPa this would be an increase of .67 kpa.
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Assuming a temperature in the neighborhood of 300 K, how much temperature change would be required, at constant volume, to achieve this pressure increase?
****
It would take about 2 degrees additional Kelvin to acheive this since it take 1 degree in Kelvin increase to move 3cm.
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The volume of the gas would change by the additional volume occupied by the water in the tube, in this case about .7 cm^3. Assuming that there are 3 liters of gas in the container, how much temperature change would be necessary to increase the gas volume by .7 cm^3?
****
.7 is .0007 Liters. this is an increase in volume of .0233 % above the original 3 Liters. An increase of temperature .0233% above 300 Kelvin is an increase of .07 degrees.
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Continue to assume a temperature near 300 K and a volume near 3 liters:
If the tube was in the completely vertical position, by how much would the position of the meniscus change as a result of a 1 degree temperature increase?
****
The menicus would move 3cm up.
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What would be the change if the tube at the position of the meniscus was perfectly horizontal? You may use the fact that the inside volume of a 10 cm length tube is .7 cm^3.
****
One degree of temp change is .33% increase. .33% increase in volume would be .0099 Liters or 9.9 mL (9.9cm^3)
9.9cm^3 divided by .7 cm^3 is 14.1 so it would take 14.1 10 cm lengths of tube to hold 9.9 mL about about 140cm of tubing.
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A what slope do you think the change in the position of the meniscus would be half as much as your last result?
****
I would estimate 30 degree slope since sine of 30 degrees is .5 or half.
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It's difficult to get much vertical change from the heat of your hands. You did manage a couple of mm. If you can manage the horizontal tube, you can get several cm from the heat of your hands.
So a change in the vertical level of the water indicates a much greater percent change in pressure than the percent change in volume due to motion in the horizontal level. So the required slope would be much less than that.
No need to revise your work on this question, though.
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Good work, except for the changes in the height of the water in the vertical tube. Check my notes. Because of some of the difficulties you've had with leaks I wouldn't ask if it wasn't important, but I will ask you for a revision on that, because those heights are very important for some later analysis. Should be pretty easy and not too time-consuming to make the revisions. If it turns out otherwise, send me some questions and I'll try to get you through that as quickly as possible.
&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes.
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Your revisions look good.
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